뒤로Equivalents, Equivalent Weight, and Normality in GOB Chemistry
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Equivalents and Equivalent Weight
Definition of Equivalents
In chemistry, equivalents are used to measure the number of H+ ions or OH- ions in acids and bases, respectively. The concept of equivalents allows chemists to compare the reactive capacities of different acids and bases.
Equivalent (Eq) of Acid: The amount of acid that contributes 1 mole of H+ ions.
Equivalent (Eq) of Base: The amount of base that contributes 1 mole of OH- ions.
Example:
1 mol HBr = 1 Eq HBr
1 mol Ca(OH)2 = 2 Eq Ca(OH)2
Calculating Equivalents
To calculate the number of equivalents of acid or base, multiply the number of moles by the number of H+ or OH- ions each molecule can donate or accept.
Type | Formula |
|---|---|
Acid Equivalent (Eq) | Eq = n × moles of acid |
Base Equivalent (Eq) | Eq = n × moles of base |
Where n = number of H+ or OH- ions per molecule.
Example: Calculating Equivalents
a) 1 mol of H3PO4: H3PO4 can donate 3 H+ ions, so n = 3. Eq = 3 × 1 mol = 3 Eq H3PO4
b) 2.7 g of ROH: Assume ROH is a monoprotic base (n = 1). Moles = 2.7 g / (molar mass of ROH) Eq = n × moles = 1 × moles of ROH
Equivalent Weight
The equivalent weight represents the mass (in grams) of an acid or base that supplies or reacts with 1 mole of H+ or OH- ions.
Formula |
|---|
Eq Weight = |
Example: Calculate the equivalent weight of H2SO4 (n = 2, since it can donate 2 H+ ions):
Molar mass of H2SO4 = 98.086 g/mol
Eq weight = g/mol
Normality (N)
Definition of Normality
Normality (N) is a measure of concentration that expresses the number of equivalents of solute per liter of solution. It is commonly used for acid-base and redox reactions.
Normality (N) =
Alternatively, Normality (N) = n × Molarity (M)
Formula | Explanation |
|---|---|
Normality (N) = | Number of equivalents per liter |
Normality (N) = n × M | n = number of H+ or OH- ions per molecule, M = molarity |
Example: Calculating Normality
a) 4.6 × 10-1 M NaOH: NaOH provides 1 OH- per molecule (n = 1). N = n × M = 1 × 0.46 = 0.46 N
b) 0.35 g of H3PO4 in 1 L: Moles = 0.35 g / 98 g/mol = 0.00357 mol n = 3 (H3PO4 can donate 3 H+) Eq = n × moles = 3 × 0.00357 = 0.0107 Eq N = Eq / 1 L = 0.0107 N
Practice Problems and Applications
Calculate mass (grams) needed for a given base equivalent: Example: 0.18 mEq of Mg(OH)2 (1.6 g needed)
Identify the acid with a given equivalent weight: Example: Which acid has an equivalent weight of 63 g? (Answer: H2SO4)
Determine volume (L) needed to prepare a solution of given normality: Example: 0.73 g of Ca(OH)2 with 1.25 N (0.9125 L needed)
Summary Table: Key Formulas
Concept | Formula |
|---|---|
Acid/Base Equivalent | Eq = n × moles |
Equivalent Weight | Eq Weight = |
Normality (N) | N = or N = n × M |
Additional info: The notes provide foundational concepts for acid-base titrations and solution preparation, which are essential for laboratory calculations in GOB Chemistry.